What is one of the maximum values of f(x) = x | vedclass

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(C) Let $f(x) = x + \sin(2x)$.First,find the derivative: $f'(x) = 1 + 2\cos(2x)$.Set $f'(x) = 0$ to find critical points: $1 + 2\cos(2x) = 0 \Rightarr

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$\frac{\pi}{3} + \frac{\sqrt{3}}{2}$

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(C) Let $f(x) = x + \sin(2x)$.First,find the derivative: $f'(x) = 1 + 2\cos(2x)$.Set $f'(x) = 0$ to find critical points: $1 + 2\cos(2x) = 0 \Rightarrow \cos(2x) = -\frac{1}{2}$.In the interval $[0, 2\pi]$,$2x$ can be $\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$.Thus,$x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.Find the second derivative: $f''(x) = -4\sin(2x)$.Check the sign of $f''(x)$ at critical points:For $x = \frac{\pi}{3}$,$f''(\frac{\pi}{3}) = -4\sin(\frac{2\pi}{3}) = -4(\frac{\sqrt{3}}{2}) = -2\sqrt{3} For $x = \frac{2\pi}{3}$,$f''(\frac{2\pi}{3}) = -4\sin(\frac{4\pi}{3}) = -4(-\frac{\sqrt{3}}{2}) = 2\sqrt{3} > 0$ (Local Minimum).Calculating the value at $x = \frac{\pi}{3}$: $f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$.

What is one of the maximum values of $f(x) = x + \sin(2x)$ in the interval $[0, 2\pi]$?

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